γλώσσες
Σελίδες
Νομικός
Characteristic polynomials and eigenvalues
for a family of tridiagonal real symmetric
matrices
Emily Gullerud
aBa Mbirika
Rita Post
January 8, 2018
“The congruence 5 ≡ 3 (mod 4) holds.”, “The polynomial λ2 − 1 has roots ±√
2.”, and
“Clearly (−2) · (−1) = 1.” –Post, aBa, and Gullerud’s embarrassments, respectively (2017)
Abstract
We explore a certain family {An}∞n=1 of n× n tridiagonal real symmetric matrices.
After deriving a three-term recurrence relation for the characteristic polynomials of
this family, we find a closed form solution. The coefficients of these characteristic poly-
nomials turn out to involve the diagonal entries of Pascal’s triangle in an attractively
inviting manner. Lastly, we explore a relation between the eigenvalues of various mem-
bers of the family. More specifically, we give a sufficient condition for when spec(Am)
is contained in spec(An).
1
Contents
1 Introduction 2
2 Preliminaries and definitions 4
3 The characteristic polynomials of An 7
3.1 A recurrence relation for fn(λ) . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3.2 The fn(λ) and their connection to Pascal’s triangle . . . . . . . . . . . . . . 8
3.3 A closed form fn(λ) for the recurrence relation . . . . . . . . . . . . . . . . . 9
3.4 Parity properties of the fn(λ) . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4 The spectrum of An 14
4.1 Roots of the first seven equations . . . . . . . . . . . . . . . . . . . . . . . . 14
4.2 A closed form for the roots of fn(λ) . . . . . . . . . . . . . . . . . . . . . . . 15
4.3 Spectral radius of An and some bounds questions . . . . . . . . . . . . . . . 18
4.4 Sufficient condition for spec(Am) ⊂ spec(An) . . . . . . . . . . . . . . . . . 19
5 Open questions 20
1 Introduction
Tridiagonal matrices arise in many science and engineering areas for example in parallel
computing, telecommunication system analysis, and in solving differential equations using
finite differences [2, 7, 4]. In particular, the eigenvalues of symmetric tridiagonal matrices
have been studied extensively starting with Golub in 1962 and since then over 40 papers
have been published on this topic as can be found on MathSciNet [5].
Tridiagonal symmetric matrices are a subclass of the class of real symmetric matrices.
A real symmetric matrix is a square matrix A with real-valued entries that are symmetric
about the diagonal; that is, A equals its transpose AT. Symmetric matrices arise naturally
in a variety of applications. Real symmetric matrices in particular enjoy the following two
properties: (1) all of their eigenvalues are real, and (2) the eigenvectors corresponding to dis-
tinct eigenvectors are orthogonal. In this paper, we investigate the family of real symmetric
2
n× n matrices of the form
An =
0 1
1. . . . . .. . . . . . 1
1 0
with ones on the superdiagonal and subdiagonal and zeroes in every other entry. The two
main results of our paper are
• MAIN RESULT 1: a closed form expression for the characteristic polynomial fn(λ)
of An and its intimate connection to the diagonals of Pascal’s triangle, and
• MAIN RESULT 2: an analysis of the spectrum spec(An) of the An matrices and,
most importantly, a proof of a closed form for the particular eigenvalues.
• MAIN RESULT 3: a sufficient condition for when spec(Am) ⊂ spec(An).
We were initially drawn to this work when noticing the first few characteristic polynomials
of our An matrices. For example, we give the first seven polynomials below.
n-value fn(λ)
1 −1λ
2 1λ2 −1
3 −1λ3 +2λ
4 1λ4 −3λ2 +1
5 −1λ5 +4λ3 −3λ
6 1λ6 −5λ4 +6λ2 −1
7 −1λ7 +6λ5 −10λ3 +4λ
Color-coding the coefficients to make the connection more transparent, we see that the four
columns of coefficients correspond directly to the the first four diagonals in Pascal’s trian-
gle. We used this observation to construct the following formula for the mth characteristic
polynomial
fm(λ) =
bm2 c∑i=0
(−1)m+i
(m− ii
)λm−2i.
We prove that this formula satisfies the three-term recurrence formula
fn(λ) = −λfn−1(λ)− fn−2(λ)
3
with initial conditions f1(λ) = −λ and f2(λ) = λ2 − 1, thereby establishing our first main
result. Our second main result was motivated by our observation that the four distinct eigen-
values of the A4 matrix are the golden ratio, its reciprocal, and their negatives. Furthermore
via Mathematica, we observed that these four eigenvalues appear in the spectrum of the An
matrices for the n-values
4, 9, 14, 19, 24, 29, 34, 39, and 44.
All of these numbers being congruent to 4 modulo 5 seemed to be no accident. This motivated
us to prove the following closed form
λs = 2 cos
(sπ
m+ 1
)for s = 1, . . . , n
for the n distinct eigenvalues of each An matrix, thereby establishing our second main result.
As an application of this, we use this closed form to give a sufficient condition for when the
eigenvalues of the matrix Am are contained in the set of eigenvalues of the matrix An, thereby
establishing our third main result.
The breakdown of the paper is as follows. In Section 2, we give some preliminaries and
definitions. In Section 3, we focus on the characteristic polynomials. In Subsection 3.1,
we derive a three-term recurrence relation for the characteristic polynomials of the family
of matrices {An}∞n=1. In Subsection 3.2, we unveil a tantalizing connection between the
coefficients of the family of characteristic polynomials {fn(λ)}∞n=1 of the matrices {An}∞n=1
and the diagonal columns of Pascal’s triangle. In Subsection 3.3, we prove that our closed
form satisfies the recurrence relation. We conclude this section by giving a parity property
of the fn(λ) in Subsection 3.4. In Section 4, we blah...
Acknowledgments
The authors thank Pizza Plus in Eau Claire, WI, where we did research at times
and enjoyed their pizza. We also thank the home of Post’s uncles Rick and Mike whose
porch hosted our research team as we further investigated this work on Memorial Day
of 2017. Finally, we thank Phoenix Park where author EmG came up with the awesome
proof of the spec containments.
2 Preliminaries and definitions
A fundamental problem in linear algebra is the determination of the values λ for which the
set of n homogeneous linear equations in n unknowns Ax = λx has a non-trivial solution.
4
The latter equation may be written in the form (A − λI)x = 0. The general theory of
simultaneous linear algebraic equations shows that there is a non-trivial solution if and only
if the matrix A − λI is singular, that is, det(A − λI) = 0. This leads to the following
fundamental definitions used in this paper.
Definition 2.1 (Eigenvalue and eigenvector). Let A ∈ Cn×n be a square matrix. An eigen-
pair of A is a pair (λ,v) ∈ C × (Cn − 0) such that Av = λv. We call λ an eigenvalue and
its corresponding nonzero vector v an eigenvector.
In particular, we often look at the set of eigenvalues of a given matrix A and the largest
element in absolute value of these eigenvalues defined as follows.
Definition 2.2 (Spectrum and spectral radius). Let A ∈ Cn×n be a square matrix. The
multiset of eigenvalues of A is called the spectrum of A and is denoted spec(A). The spectral
radius of A is denoted ρ(A) and is given by
ρ(A) = max{|λ| : λ ∈ spec(A)}.
The matrices we care about in this paper are a subset of a wider class of symmetric
matrices called Toeplitz matrices. A Toeplitz matrix is a matrix of the form
a0 a1 a2 . . . . . . an−1
a1 a0 a1. . .
...
a2 a1. . . . . . . . .
......
. . . . . . . . . a1 a2...
. . . a1 a0 a1
an−1 . . . . . . a2 a1 a0
.
In particular, we focus on tridiagonal symmetric matrices. We give the specific definitions
below.
Definition 2.3 (Tridiagonal matrix). A tridiagonal symmetric matrix is a Toeplitz matrix
in which all entries not lying on the diagonal or superdiagonal or subdiagonal are zero.
We limit our perspective by considering the tridiagonal matrices of the following form.
Definition 2.4 (The An matrices). Let An be an n × n tridiagonal symmetric matrix in
which the diagonal entries are zero, and the superdiagonal and subdiagonal entries are all
one.
5
Example 2.5. Here are the An matrices for n = 1, . . . , 4.
A1 =[0]
A2 =
[0 1
1 0
]A3 =
0 1 0
1 0 1
0 1 0
A4 =
0 1 0 0
1 0 1 0
0 1 0 1
0 0 1 0
A typical method of finding the eigenvalues of a square matrix is by calculating the roots
of what is called the characteristic polynomial of the matrix.
Definition 2.6 (The characteristic polynomial fn(λ)). Given the matrix An. The charac-
teristic polynomial fn(λ) is given by
fn(λ) = det(An − λIn).
If we set fn(λ) = 0, then clearly we have that the n zeroes of this equation give the
spectrum of An.
Example 2.7. Consider the following real symmetric matrix A4 and the corresponding
determinant |A4 − λI4|.
A4 =
0 1 0 0
1 0 1 0
0 1 0 1
0 0 1 0
|A4 − λI4| =
∣∣∣∣∣∣∣∣∣∣−λ 1 0 0
1 −λ 1 0
0 1 −λ 1
0 0 1 −λ
∣∣∣∣∣∣∣∣∣∣= λ4 − 3λ2 + 1.
Thus the characteristic polynomial is f4(λ) = λ4 − 3λ2 + 1. The roots of this polynomial
yield the four distinct eigenvalues
λ1 =1 +√
5
2= φ ≈ 1.61803 λ3 =
−1 +√
5
2=
1
φ≈ .61803
λ2 =1−√
5
2≈ −.61803 λ4 =
−1−√
5
2≈ −1.61803,
where φ is the golden ratio. Moreover in Section 4.4, we find an infinite set of n values for
which these four eigenvalues in the n = 4 case are contained in the spectrum of larger An
matrices.
6
3 The characteristic polynomials of An
In this secction we find the family of characteristic polynomials {fn(λ)}∞n=1 corresponding to
our family of matrices. For each n, the coefficients of the polynomial fn(λ) reveal themselves
to be exactly the (n + 1)th diagonal of Pascal’s triangle as denoted in 3.1. We use this
observation to find a closed form expression for fn(λ) for all n. To confirm this expression
is indeed correct, we show that it satisfies the easily derived three-term recurrence relation
that the sequence (fn(λ))∞n=1 must follow.
3.1 A recurrence relation for fn(λ)
In this subsection, we show that the determinants of the matrix An − λIn, that is, the
characteristic polynomials fn(λ), satisfy the following three-term recurrence relation
fn(λ) = −λfn−1(λ)− fn−2(λ)
with initial conditions f1(λ) = −λ and f2(λ) = λ2 − 1. First we consider the n × n matrix
An−λIn and note that bottom-right (n− 1)× (n− 1) submatrix is An−1−λIn−1, and hence
the minor given by deletion of the first row and column is the characteristic polynomial
fn−1(λ). This fact is pivotal in deriving the recurrence relation fn(λ).
An − λIn =
−λ 1 0 · · · 0
1 −λ 1. . . 0
0 1 −λ . . ....
.... . . . . . . . . 1
0 0 · · · 1 −λ
=
−λ 1 0 . . . 0
1
0 An−1−λIn−1...
0
Computing the determinant by cofactor expansion along the top row of An− λIn, we derive
the following.
fn(λ) = |An − λIn|
7
= −λ · |An−1 − λIn−1| − 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣
1 1 0 · · · 0
0 −λ 1. . . 0
0 1 −λ . . ....
.... . . . . . . . . 1
0 0 · · · 1 −λ
∣∣∣∣∣∣∣∣∣∣∣∣∣∣
= −λ · fn−1(λ)− 1
1 1 0 . . . 0
0... An−2−λIn−2...
0
∣∣∣∣∣∣∣∣∣∣∣∣∣∣
∣∣∣∣∣∣∣∣∣∣∣∣∣∣= −λ · fn−1(λ)− fn−2(λ).
3.2 The fn(λ) and their connection to Pascal’s triangle
If we compute the characteristic polynomials fn(λ) using the determinant of An − λIn, we
get the following table of polynomials for n = 1, . . . , 12.
n-value fn(λ)
1 −1λ
2 1λ2 −1
3 −1λ3 +2λ
4 1λ4 −3λ2 +1
5 −1λ5 +4λ3 −3λ
6 1λ6 −5λ4 +6λ2 −1
7 −1λ7 +6λ5 −10λ3 +4λ
8 1λ8 −7λ6 +15λ4 −10λ2 +1
9 −1λ9 +8λ7 −21λ5 +20λ3 −5λ
10 1λ10 −9λ8 +28λ6 −35λ4 +15λ2 −1
11 −1λ11 +10λ9 −36λ7 +56λ5 −35λ3 +6λ
12 1λ12 −11λ10 +45λ8 −84λ6 +70λ4 −21λ2 +1
Upon examination of the coefficients of these 12 polynomials, it becomes immediately appar-
8
Figure 3.1: Pascal’s triangle
ent that the binomial coefficients in the diagonal columns of Pascal’s triangle are intimately
involved in the coefficients of the fn(λ). In Figure 3.1 we give the first 12 rows of Pascal’s tri-
angle, carefully coding the diagonal columns, to match the table of the 12 fn(λ) polynomials
with their corresponding colors matching those in Pascal’s triangle.
3.3 A closed form fn(λ) for the recurrence relation
To arrive at our closed form for the solution to the recurrence relation, we use our observation
of the intimate connection with Pascal’s triangle and the coefficients of the individual fn(λ)
as n varies. For instance, the 1st, 2nd, etc. coefficients of each fn(λ) coincide directly with
entries in the 1st, 2nd, etc.diagonals of Pascal’s triangle in the particular manner depicted in
Figure 3.1. We highlight this connection to make this relationship apparent.
By careful analysis of the connection between the fn(λ) polynomials and Pascal’s triangle,
we prove our main result that a closed form for the recurrence relation in Subsection 3.1 is
given by
fm(λ) =
bm2 c∑i=0
(−1)m+i
(m− ii
)λm−2i. (1)
9
In actual practice, it is more helpful to consider the equation above for m even and m odd.
The closed form when the index m is even is
f2k(λ) =k∑i=0
(−1)i(
2k − ii
)λ2k−2i. (2)
And the closed form when the index m is odd is
f2k+1(λ) =k∑i=0
(−1)i+1
((2k + 1)− i
i
)λ(2k+1)−2i. (3)
Before we prove that the recurrence relation fn(λ) = −λfn−1(λ)− fn−2(λ) is satisfied by
these closed forms above, we give a motivating example for when n is even. The case for
when n is odd is similar.
Example 3.1. (An even case example) Set n = 10. According to the recurrence relation,
f10(λ) = −λf9(λ)− f8(λ). (?)
By Equation (2), the left hand side of (?) is
f10(λ) =
5∑i=0
(−1)i(
10− ii
)λ10−2i =
(10
0
)λ10 −
(9
1
)λ8 +
(8
2
)λ6 −
(7
3
)λ4 +
(6
4
)λ2 −
(5
5
).
By Equations (2) and (3), the right hand side of (?) is
−λf9(λ)− f8(λ) = −λ4∑
i=0
(−1)i+1
(9− ii
)λ9−2i −
4∑i=0
(−1)i(
8− ii
)λ8−2i
= −λ[−(90
)λ9 +
(81
)λ7 −
(72
)λ5 +
(63
)λ3 −
(54
)λ
]−[(8
0
)λ8 −
(71
)λ6 +
(62
)λ4 −
(53
)λ2 +
(44
)]=(90
)λ10 −
(81
)λ8 +
(72
)λ6 −
(63
)λ4 +
(54
)λ2 −
(80
)λ8 +
(71
)λ6 −
(62
)λ4 +
(53
)λ2 −
(44
)=(90
)λ10 −
[(81
)+(80
)]λ8 +
[(72
)+(71
)]λ6 −
[(63
)+(62
)]λ4 +
[(54
)+(53
)]λ2 −
(44
)=(100
)λ10 −
(91
)λ8 +
(82
)λ6 −
(73
)λ4 +
(64
)λ2 −
(55
)= f10(λ).
The second to last line above utilizes the fact that(90
)=(100
)and
(44
)=(55
)trivially, as well
as the property of Pascal’s triangle where(n
r
)+
(n
r + 1
)=
(n+ 1
r + 1
). (4)
We now state and prove our main result.
10
Theorem 3.2. Consider the recurrence relation fn(λ) = −λfn−1(λ) − fn−2(λ) with initial
conditions f1(λ) = −λ and f2(λ) = λ2 − 1. A closed form for this recurrence relation is
given by
fm(λ) =
bm2 c∑i=0
(−1)m+i
(m− ii
)λm−2i.
Proof. Consider the recurrence relation defined by
fn(λ) = −λfn−1(λ)− fn−2(λ).
It is readily verified that Equation (1) yields the initial conditions when n = 1, 2, so it suffices
to show that Equation (1) satisfies the recurrence relation. We divide this into two separate
cases depending on the parity of n.
CASE 1: (n is even)
Since n ∈ 2N then n = 2k for some k ∈ N. Hence we want to show that the following
equation is satisfied by our appropriate choice of closed forms Equations (2) and (3) as
needed depending on the parity of each of the indices.
f2k(λ) = −λf2k−1(λ)− f2k−2(λ) (†)
We start with the right hand side of (†). For the ease of the reader, the expressions in red
indicate the change from one line to the next.
−λf2k−1(λ)− f2k−2(λ)
= −λk−1∑i=0
(−1)i+1
((2k − 1)− i
i
)λ(2k−1)−2i −
k−1∑i=0
(−1)i(
(2k − 2)− ii
)λ(2k−2)−2i (5)
=
k−1∑i=0
(−1)i(
(2k − 1)− ii
)λ2k−2i +
k−1∑i=0
(−1)i+1
((2k − 2)− i
i
)λ(2k−2)−2i
=
(2k − 1
0
)λ2k +
k−1∑i=1
(−1)i(
(2k − 1)− ii
)λ2k−2i
+
k−2∑i=0
(−1)i+1
((2k − 2)− i
i
)λ(2k−2)−2i+ (−1)k
(k − 1
k − 1
)λ0
=
(2k − 1
0
)λ2k +
k−2∑j=0
(−1)j+1
((2k − 1)− (j + 1)
j + 1
)λ2k−2(j+1)
+
k−2∑j=0
(−1)j+1
((2k − 2)− j
j
)λ(2k−2)−2j + (−1)k
(k − 1
k − 1
)λ0
(6)
11
=
(2k − 1
0
)λ2k +
k−2∑j=0
(−1)j+1
[((2k − 1)− (j + 1)
j + 1
)+
((2k − 2)− j
j
)]λ2k−2−2j
+ (−1)k(k − 1
k − 1
)λ0
=
(2k − 1
0
)λ2k +
k−2∑j=0
(−1)j+1
((2k − 2− j) + 1
j + 1
)λ2k−2−2j + (−1)k
(k − 1
k − 1
)λ0 (7)
=
(2k
0
)λ2k +
k−1∑i=1
(−1)i(
2k − ii
)λ2k−2i + (−1)k
(2k − kk
)λ2k−2k (8)
=
k∑i=0
(−1)i(
2k − ii
)λ2k−2i
= f2k(λ) (9)
Line (6) holds by the closed forms given by Equations (2) and (3). Line (7) holds by
taking i = j + 1 for the first summation and i = j for the second summation. Line (8) holds
by Equation (4). Line (9) holds by taking i = j + 1 and realizing that(2k−10
)=(2k0
)and(
k−1k−1
)=(kk
). Line (10) holds by the closed form given by Equation (2). Therefore the closed
forms satisfy the recurrence relation when n is even.
CASE 2: (n is odd)
Since n ∈ 2N+ 1 then n = 2k+ 1 for some k ∈ N. Hence we want to show that the following
equation is satisfied by our appropriate choice of closed forms Equations (2) and (3) as
needed depending on the parity of each of the indices.
f2k+1(λ) = −λf2k(λ)− f2k−1(λ) (††)
We start with the right hand side of (††):
−λf2k(λ)− f2k−1(λ)
= −λk∑
i=0
(−1)i(
2k − ii
)λ2k−2i −
k−1∑i=0
(−1)i+1
((2k − 1)− i
i
)λ(2k−1)−2i (10)
=
k∑i=0
(−1)i+1
(2k − ii
)λ2k−2i+1 +
k−1∑i=0
(−1)i(
(2k − 1)− ii
)λ(2k−1)−2i
12
= −(
2k
0
)λ2k+1 +
k∑i=1
(−1)i+1
(2k − ii
)λ2k−2i+1
+
k∑j=1
(−1)j−1
((2k − 1)− (j − 1)
j − 1
)λ(2k−1)−2(j−1)
(11)
= −(
2k
0
)λ2k+1 +
k∑j=1
(−1)j+1
(2k − jj
)λ2k−2j+1 +
k∑j=1
(−1)j+1
(2k − jj − 1
)λ2k−2j+1 (12)
= −(
2k
0
)λ2k+1 +
k∑j=1
(−1)j+1
[(2k − jj
)+
(2k − jj − 1
)]λ2k−2j+1
= −(
2k + 1
0
)λ2k+1 +
k∑j=1
(−1)j+1
((2k + 1)− j
j
)λ(2k+1)−2j (13)
=
k∑j=0
(−1)j+1
((2k + 1)− j
j
)λ(2k+1)−2j
= f2k+1(λ) (14)
Line (12) holds by the closed forms given by Equations (2) and (3). Line (13) holds
by taking i = j − 1 for the second summation. Line (14) holds by taking i = j for the
first summation. Line (15) holds by Equation (4) and realizing that(2k0
)=(2k+10
). Line
(16) holds by the closed form given by Equation (3). Therefore the closed forms satisfy the
recurrence relation when n is odd.
Hence Equation (1) satisfies the recurrence relation.
3.4 Parity properties of the fn(λ)
Observing the graphs of various characteristic functions fn(λ) in Subsection 4.1, it appears
that the functions of even index are symmetric about the y-axis (the hallmark trait of an
even function), while the functions of odd index are symmetric about the origin—that is,
the graph remains unchanged after 180 degree rotation about the origin (the hallmark trait
of an odd function). This observation prompted the following theorem.
Theorem 3.3. The characteristic equation fn(λ) is an even (respectively, odd) function
when n is even (respectively, odd).
Proof. It suffices to show
• Claim 1: f2k(−λ) = f2k(λ), and
13
• Claim 2: f2k+1(−λ) = −f2k+1(λ).
Using our closed form Equation (2) when the index is even, we see that Claim 1 holds since
(−λ)2k−2i =((−λ)2
)k−i= (λ2)k−i
= λ2k−2i.
(15)
And using our closed form Equation (3) when the index is odd, we see that Claim 2 holds
since
(−λ)(2k+1)−2i = (−λ)(−λ)2k−2i
= (−λ)λ2k−2i by the even case result (15)
= −(λ)(2k+1)−2i.
4 The spectrum of An
aBa: [Put some introductory words here on why people care about eigenvalues.]
4.1 Roots of the first seven equations
Here are the exact roots of the first seven characteristic functions.
n-value exact roots of fn(λ)
1 λ = 0
2 λ = 1 or λ = −13 λ = 0 or λ =
√2 or λ = −
√2
4 λ = 12
(−√5− 1
)or λ = 1
2
(√5− 1
)or λ = 1
2
(1−√5)
or λ = 12
(√5 + 1
)5 λ = 0 or λ =
√3 or λ = −
√3 or λ = −1 or λ = 1
6 λ = 13
(3
√72
(3i√3 + 1
)+ 72/3
3√
12 (3i√3+1)
− 1
)or λ = − 1
6
(1− i
√3)
3
√72
(3i√3 + 1
)− 1
3− 72/3(i
√3+1)
3 22/33√
3i√3+1
or
λ = − 16
(i√3 + 1
)3
√72
(3i√3 + 1
)− 1
3− 72/3(1−i
√3)
3 22/33√
3i√3+1
or λ = 13
(3
√72
(3i√3− 1
)+ 72/3
3√
12 (3i√3−1)
+ 1
)or
λ = − 16
(1− i
√3)
3
√72
(3i√3− 1
)+ 1
3− 72/3(i
√3+1)
3 22/33√
3i√3−1
or λ = − 16
(i√3 + 1
)3
√72
(3i√3− 1
)+ 1
3− 72/3(1−i
√3)
3 22/33√
3i√3−1
7 λ = 0 or λ =√
2−√2 or λ = −
√2−√2 or λ =
√√2 + 2 or λ = −
√√2 + 2 or λ =
√2 or λ = −
√2
14
n-value roots (up to 5 decimal places) of fn(λ) in increasing order
1 λ = 0
2 λ = −1 or λ = 1
3 λ = −1.41421 or λ = 0 or λ = 1.41421
4 λ = −1.61803 or λ = −0.618034 or λ = 0.618034 or λ = 1.61803
5 λ = −1.73205 or λ = −1 or λ = 0 or λ = 1 or λ = 1.73205
6 λ = −1.80194 or λ = −1.24698 or λ = −0.445042 or λ = 0.445042 or λ = 1.24698 or λ = 1.80194
7 λ = −1.84776 or λ = −1.41421 or λ = −0.765367 or λ = 0 or λ = 0.765367 or
λ = 1.41421 or λ = 1.84776
In Figure 4.1, we plot the characteristic equations fn(λ) for n = 2, . . . , 7. Notice that since
the odd-index equations have no constant term, the graphs of those equations necessarily
pass through the origin.
Figure 4.1: Plots of fn(λ) for n = 2, . . . , 7
4.2 A closed form for the roots of fn(λ)
Though many sources going back as far as 1965 are credited in later papers as giving closed
forms for the eigenvalues of tridiagonal symmetric matrices, none of these cited journals and
books actually provide a proof [1, 3, 6]. However in his 1978 book, Smith provides a proof for
15
the closed forms for eigenvalues of tridiagonal (but not necessarily) symmetric matrices that
have the values a, b, c ∈ C in the diagonal, superdiagonal, and subdiagonal, respectively [8].
For these matrix values, he computes the n eigenvalues to be the formula
λs = a+ 2√bc cos
sπ
n+ 1for s = 1, . . . , n.
Below we modify the proof to give a closed form expression for the eigenvalues of the An
matrices.
Proposition 4.1. The n distinct eigenvalues of An are
λs = 2 cos
(sπ
n+ 1
)for s = 1, . . . , n.
Proof. Let λ represent an eigenvalue of An with corresponding eigenvector ~v. Consider the
equation An~v = λ~v. This can be rewritten as (An − λIn)~v = ~0. Then
(An − λIn)~v =
−λ 1
1. . . . . .. . . . . . 1
1 −λ
~v1
~v2...
~vn
=
0
0...
0
.The resulting system of equations is as follows:
−λ~v1 + ~v2 = 0
~v1 − λ~v2 + ~v3 = 0
......
~vj−1 − λ~vj + ~vj+1 = 0
......
~vn−1 − λ~vn = 0
In general, a single equation can be defined by
~vj−1 − λ~vj + ~vj+1 = 0 for j = 1, . . . , n (16)
where ~v0 = 0 and ~vn+1 = 0. Let a solution to this equation be ~vj = Amj for arbitrary
non-zero constants A and m. Substituting this solution into Equation (16) shows that m is
a root of
m2 − λm+ 1. (17)
16
Since Equation (17) is a quadratic equation, there are two, in our case real, roots. Hence
the general solution of Equation (16) is
~vj = Bmj1 + Cmj
2 (18)
where B and C are arbitrary non-zero constants and m1 and m2 are the two roots of
Equation (17). Since ~v0 = 0 and ~vn+1 = 0, by Equation (17), we get B + C = 0 and
Bmn+11 + Cmn+1
2 = 0. Substituting the former equation into the latter results in(m1
m2
)n+1
= 1 = e2πis for s = 1, . . . , n.
Hence it follows thatm1
m2
= e2πis/(n+1). (19)
Since Equation (17) is a quadratic equation, the product of the roots is given by
m1m2 = 1. (20)
Then using Equation (19),
m1
m2
= e2πis/(n+1)
⇒ m1 = m2e2πis/(n+1)
⇒ m21 = m1m2e
2πis/(n+1)
⇒ m21 = e2πis/(n+1) by Equation (20)
⇒ m1 = eπis/(n+1).
Through a similar argument it can be found that
m2 = e−πis/(n+1).
Again since Equation (17) is a quadratic equation, the sum of the roots is given by
m1 +m2 = λ.
It follows that
λ = m1 +m2
= eπis/(n+1) + e−πis/(n+1)
= 2 cos
(πs
n+ 1
).
Therefore λs = 2 cos(πsn+1
)for s = 1, . . . , n.
17
4.3 Spectral radius of An and some bounds questions
Much work has been done to study the bounds of the eigenvalues of real symmetric matrices.
Methods discovered in the mid-nineteenth century reduce the original matrix to a tridiagonal
matrix whose eigenvalues are the same as those of the original matrix. Exploiting this idea,
Golub in 1962 determined lower bounds on tridiagonal matrices of the form
a0 b1 0 . . . . . . 0
b1 a2 b2. . .
...
0 b2. . . . . . . . .
......
. . . . . . . . . bn−2 0...
. . . bn−2 an−1 bn−1
0 . . . . . . 0 bn−1 an
.
Proposition 4.2 (Golub [5, Corollary 1.1]). Let A be an n × n matrix with real entries
aij = ai for i = j, aij = bm for |i− j| = 1 where m = min(i, j), and aij = 0 otherwise. Then
the interval [ak − σk, ak + σk] where σ2k = b2k + b2k−1 contains at least one eigenvalue.
Using the proposition above, we can apply that to the tridiagonal matrices An. We easily
yield an interval in which the lower bound (in absolute value) of the eigenvalues of An is
guaranteed to occur and also be achieved.
Corollary 4.3. Given the matrix An, there exists an eigenvalue λ such that |λ| ≤ 1.
Proof. Given An, then in the language of Proposition 4.2, we have ai = 0 and bi = 1 for each
i. Thus the σk are calculated as follows
σk =
1, if k = 1√
2, if 2 ≤ k ≤ n− 1
1, if k = n
.
By Proposition 4.2, an eigenvalue is guaranteed to exist in the the interval [ak−σk, ak +σk].
In particular for k = 1 the result follows.
So we have a firm interval where the lower bound (in absolute value) will contain an
eigenvalue of An. However, the question still remains on precise values for the upper bound
of the spectrum of An, namely the spectral radius ρ(An) (see Question 5.2).
18
4.4 Sufficient condition for spec(Am) ⊂ spec(An)
Recall in Section 3, we gave the characteristic equation when n = 7 as follows
f7(λ) = −λ7 + 6λ5 − 10λ3 + 4λ.
In the table in Subsection 4.1, for n = 7 the equation above has the following seven roots
λ = 0
λ = ±√
2
λ = ±√
2−√
2
λ = ±√√
2 + 2.
It is interesting to note that these exact seven roots appear as eigenvalues in a higher-degree
characteristic equation. For example, when n = 15, the characteristic equation is
f15(λ) = −λ15 + 14λ13 − 78λ11 + 220λ9 − 330λ7 + 252λ5 − 84λ3 + 8λ.
The 15 distinct roots are the following
λ = 0
λ = ±√
2
λ = ±√
2−√
2
λ = ±√√
2 + 2
λ = ±√
2−√
2−√
2
λ = ±√√
2−√
2 + 2
λ = ±√
2−√√
2 + 2
λ = ±√√√
2 + 2 + 2.
This phenomenon is no coincidence. In fact the following theorem provides a way to
predict the values m < n for which a complete set of eigenvalues of fm(λ) will be contained
in the set of eigenvalues for fn(λ).
Theorem 4.4. Fix m ∈ N. For any n ∈ N such that m < n and n ≡ m (mod m + 1), it
follows that spec(Am) ⊂ spec(An).
Proof. Let m,n ∈ N such that m < n and n ≡ m (mod m+ 1). It follows that
n = (m+ 1)k +m for some k ∈ N.
19
This can be rewritten as
n = m(k + 1) + k. (21)
Now consider spec(Am) and spec(An). By Proposition 4.1, these are defined as follows:
spec(Am) = {λr | λr = 2 cos(
rπm+1
)}mr=1
spec(An) = {λs | λs = 2 cos(sπn+1
)}ns=1.
We rewrite λs in terms of m and k:
λs = 2 cos
(sπ
n+ 1
)for s = 1, 2, . . . , n
= 2 cos
(sπ
m(k + 1) + k + 1
)for s = 1, 2, . . . ,m(k + 1) + k by Equation (21)
= 2 cos
(sπ
mk +m+ k + 1
)= 2 cos
(sπ
(m+ 1)(k + 1)
)= 2 cos
((s
k+1
)π
m+ 1
).
Recall λr = 2 cos(
rπm+1
). Hence λr = λs when r = s
k+1.
Since s ∈ S = {1, 2, . . . ,m(k + 1) + k}, the set T = {k + 1, 2(k + 1), . . . ,m(k + 1)} is a
subset of S. Notice for each s ∈ T , there exists a unique r ∈ {1, 2, . . . ,m} under the equality
r = sk+1
. Thus for all λr ∈ spec(Am), there exists a λs ∈ spec(An) such that λr = λs. Then
λr ∈ spec(An) for all r ∈ {1, 2, . . . ,m}. Therefore spec(Am) ⊂ spec(An).
Remark 4.5. In the introduction we noted our observation that the golden ratio, its re-
ciprocal, and their negatives arise as eigenvalues of An for the n values 4, 9, 14, 19, 24,
29, 34, 39, and 44 via Mathematica. In Example 2.7, we computed the four eigenvalues
of A4. As an application of Theorem 4.4, if we let m = 4, then it is evident that the
spec(A4) ⊆ spec(A4+5k) for all k ∈ N, thus confirming our observation.
5 Open questions
Question 5.1. In the plot in Figure 4.1 it appears that the maximum and minimum values
for the characteristic equations lie on a hyperbola. For example, here is a plot of f10(λ) and
f20(λ).
20
Do the max and min values of the sequence {fn(λ)}∞n=1 lie on a particular hyperbola? And
if so, can we find an exact formula for this hyperbola?
Question 5.2. Analyzing the plots in Figure 4.1 and roots of the sequence of fn(λ) functions
as the values of n increase from 1 to 46, the largest roots slowly approach 2. Moreover this
growth appears to logarithmically increase as n increases. For example, when n = 100 the
largest eigenvalue is approximately 2.49734, and when n = 1000 the largest eigenvalue is
approximately 7.43481. Are there arbitrarily large eigenvalues (in absolute value), or do
these eigenvalues approach some bound as n tends to infinity?
aBa: [QUESTION: Am I using logarithmic growth correctly? Maybe we should plot the
max values in the y-axis with n-values in the x-axis, and see what the plot looks like?? Use
Mathematica.]
aBa: [Look up Wikipedia page on spectral radius. Apparently, the spectral radius ρ(A) of a
matrix A is has an upper bound using a function involving the matrix norm.]
References
[1] M. Abromovitz and I. A. Stegun, Handbook of Mathematical Functions, Dover, New York
(1965).
[2] I. Bar-On, Interlacing properties of tridiagonal symmetric matrices with applications
to parallel computing, SIAM Journal on Matrix Analysis and Applications, 17 (1996),
p. 548–562.
21
[3] S. Barnett, Matrices, Methods and Applications, Oxford University Press, Oxford (1990).
[4] C. Cryer, The numerical solution of boundary value problems for second order functional
differential equations by finite differences, Numerische Mathematik, 20 (1972), p. 288–
299.
[5] G. Golub, Bounds for eigenvalues of tridiagonal symmetric matrices computed by the
LR method, Mathematics of Computation 16 (1962), p. 428–447.
[6] W. Kahan and J. Varah, Two working algorithms for the eigenvalues of a symmetric
tridiagonal matrix, Technical Report CS43, Computer Science Department, Stanford
University (1966), 29pp.
[7] L. Lu and W. Sun, The minimal eigenvalues of a class of block-tridiagonal matrices,
IEEE Trans. Inform. Theory 43 (1997), p. 787–791.
[8] G.D. Smith, Numerical Solution of Partial Differential Equation: Finite Difference Meth-
ods, 2nd ed., Clarendon Press, Oxford University Press, New York (1978).
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